um help!!
(x^2-64/5x) * (10x^2/x+8)
2006-09-07 04:21:57 · 3 answers · asked by yupthatzme27 2 in Education & Reference ➔ Homework Help
umm, first guy that answeres, not quite what i was looking for seems logical, just not what i learned
2006-09-07 04:54:30 · update #1
u have to figure out a way to CANCEL stuff out!! :) so:
(x^2-64)/5x * (10x^2/x+8)
(x-8)(x+8) * (5x)(2x)
__________________
5x * (x+8)
cancel out the (x+8) and the 5x on top and bottom
(x-8)(2x)
= 2x^2 - 16x
that's my answer! good luck~
2006-09-07 04:59:23 · answer #1 · answered by sasmallworld 6 · 0⤊ 0⤋
(x^2 - 64/5x) * (10x^2/x+8)
Step 1: Create a common fraction from the first part.
x^2 - 64/5x
x^2 * 5x/5x - 64/5x
(5x^3 - 64)/5x
Step 2: Multiply the fractions.
(5x^3 - 64)/5x * 10x^2 / x+8
(5x^3 - 64) * 10x^2 / (5x * (x+8)
50x^5 - 640x^2 / 5x^2 + 40
Step 3: Simplify:
50x^5 - 640x^2 / 5x^2 + 40
10x^5 - 128x^2 / x^2 + 8 (Solution!)
2006-09-07 11:49:08 · answer #2 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
The second answer is the easiest way. Your just pulling out a common factor (simplifying) then using FOIL method.
2006-09-07 12:05:24 · answer #3 · answered by BOB W 3 · 0⤊ 0⤋